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Nuetrik [128]
2 years ago
8

If two parallel conductors are both free to move and are carrying current in the same direction, what would happen if the curren

t flow in each wire was reversed at the same time?A. The wires would repel, then attract each other. B. The wires would repel each other as they had before C. The wires would attract, then repel each other. D. The wires would remain attracted to each other.
Physics
1 answer:
pishuonlain [190]2 years ago
8 0

The wires would remain attracted to each other.

Option D.

Explanation:

It is known that magnetic flux will be generated in conductors with varying emf. So when current is flowing in two parallel conductors, the magnetic flux will be generated in those wires. If the current is flowing in same direction in both the wires, then the magnetic flux will be generated towards inside and outside the wires. Thus, the wire will get attracted to each other till the time the current is flowing in the same direction in both the wires. So if the current flow in each wire was reversed at the same time, then the wire would remain attracted to each other.

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Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves f
stealth61 [152]

Answer:

20.96 h

Explanation:

The perimeter of the track is 2*pi*r = 20pi miles

In 10 hours, car B would have moved 20miles. So, when Car A leaves from point X, car B is 20pi - 20 miles from point X counter-clockwise and car A.

From here, we can express the distance of A from X like this:

xa = 3t

And the distance of B would be:

xb = 20pi - 20 - 2t

The time t where they would passed each other and put  12 miles between them would be the one where xa - xb is equal to 12:

xa - xb = 12

3t - (20pi - 20 - 2t) = 12

5t = 20 pi - 8

t = (20pi - 8)/5 = 10.96 h

Remember to add this value to the 10 hours car B had already been racing:

t = 20.96h

4 0
3 years ago
HELPP!! what is not true about mitosis??
Andreyy89

Answer:

I think its A

Explanation:

7 0
3 years ago
Read 2 more answers
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragst
disa [49]

Answer:

a)  Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) The displacement of the dragster at the end of 1.8 s = 68.04 m

d) The displacement of the dragster at the end of 3.6 s = 272.16 m

Explanation:

a) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

  v = u + at

  v  = 0 + 42 x 1.8 = 75.6 m/s

Final velocity of the dragster at the end of 1.8 s = 75.6 m/s

b) We have equation of motion v = u + at

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

  v = u + at

  v  = 0 + 42 x 3.6 = 75.6 m/s

Final velocity of the dragster at the end of 3.6 s = 151.2 m/s

c) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 1.8 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 1.8 + 0.5 x 42 x 1.8²

    s = 68.04 m

The displacement of the dragster at the end of 1.8 s = 68.04 m

d) We have equation of motion s= ut + 0.5 at²

  Initial velocity, u =  0 m/s

 Acceleration , a = 42 m/s²

 Time = 3.6 s    

Substituting

   s= ut + 0.5 at²

    s = 0 x 3.6 + 0.5 x 42 x 3.6²

    s = 272.16 m

The displacement of the dragster at the end of 3.6 s = 272.16 m

3 0
2 years ago
What is the numerical value of the slope of the line (without units)?<br>0.083<br>24<br>8.3<br>12
tankabanditka [31]
As close as I can read it, it appears to be

                 1/12  gram/second

           (0.08333... gm/sec)
4 0
3 years ago
The magnitude of the tidal force between the International Space Station (ISS) and a nearby astronaut on a spacewalk is approxim
vovikov84 [41]

Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

      F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

      F = 2 6.67 10⁻¹¹ 135  5.98 10²⁴ 13 / (6.79 10⁶)³

      F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

4 0
2 years ago
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